Question: Let $a(x)=-2x^7+x^5+3x^4+2x^2-7$, and $b(x)=x^4$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's rewrite the fraction to cancel common factors: $ \begin{aligned} \dfrac{a(x)}{b(x)}=\dfrac{-2x^7+x^5+3x^4+2x^2-7}{x^4}&=\dfrac{-2 {x^7}+ {x^5}+3 {x^4}}{ {x^4}}+\dfrac{2x^2-7}{x^4}\\\\ &={-2x^3+x+3}+\dfrac{{2x^2-7}}{x^4}\\\\ &={q(x)} + \dfrac{{r(x)}}{b(x)}\end{aligned}$ Since the degree of ${2x^2-7}$ is less than the degree of $x^4$, it follows that ${r(x)}={2x^2-7}$, and ${q(x)}={-2x^3+x+3}$. To conclude, $q(x)=-2x^3+x+3$ $r(x)=2x^2-7$ [Is there another way of doing this?]